A proton collides with a stationary deuteron to form a `.^3He` nucleus. For this reaction to take place , the proton must have a minimum kinetic energy `K_0` . If instead , a deuteron collides with a stationary proton to make a `.^3He` nucleus , then it must have minimum kinetic energy equal to

Options

A. `2K_0`
B. `1.5K_0`
C. `K_0`
D. `K_0/2`

Answer

To solve the problem, we need to analyze the two scenarios involving the collisions of protons and deuterons to form a helium-3 nucleus. We will use the principles of conservation of energy and momentum.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
- In the first scenario, a proton (mass = m) collides with a stationary deuteron (mass = 2m) to form a helium-3 nucleus (mass = 3m).
- In the second scenario, a deuteron collides with a stationary proton.

2. **Minimum Kinetic Energy for the Proton**:
- Let the minimum kinetic energy of the proton required for the first reaction be \( K_0 \).
- The energy conservation equation before and after the reaction can be written as:
\[ K_0 + 0 = \frac{1}{2} (3m) v_{He}^2 \]
- Here, \( v_{He} \) is the velocity of the helium nucleus after the collision.

3. **Velocity of Helium Nucleus**:
- From momentum conservation, the momentum before the collision (only the proton is moving) must equal the momentum after the collision:
\[ mv_p = 3m v_{He} \]
- Thus, we can express \( v_{He} \) in terms of \( v_p \):
\[ v_{He} = \frac{v_p}{3} \]

4. **Substituting Velocity into Energy Equation**:
- Substituting \( v_{He} \) into the energy conservation equation:
\[ K_0 = \frac{1}{2} (3m) \left(\frac{v_p}{3}\right)^2 \]
- Simplifying this gives:
\[ K_0 = \frac{1}{2} (3m) \left(\frac{v_p^2}{9}\right) = \frac{1}{6} mv_p^2 \]

5. **Second Scenario with Deuteron**:
- Now consider the second scenario where a deuteron collides with a stationary proton.
- Let the kinetic energy of the deuteron be \( K_D \).
- The energy conservation equation for this scenario is:
\[ K_D + 0 = \frac{1}{2} (3m) v_{He}^2 \]

6. **Momentum Conservation for Deuteron**:
- The momentum conservation equation gives:
\[ 2m v_D = 3m v_{He} \]
- Thus, we can express \( v_{He} \) in terms of \( v_D \):
\[ v_{He} = \frac{2}{3} v_D \]

7. **Substituting Velocity into Energy Equation**:
- Substituting \( v_{He} \) into the energy conservation equation:
\[ K_D = \frac{1}{2} (3m) \left(\frac{2}{3} v_D\right)^2 \]
- Simplifying this gives:
\[ K_D = \frac{1}{2} (3m) \left(\frac{4}{9} v_D^2\right) = \frac{2}{3} mv_D^2 \]

8. **Relating \( K_D \) to \( K_0 \)**:
- From the first scenario, we know \( K_0 = \frac{1}{6} mv_p^2 \).
- Since the minimum kinetic energy required for the deuteron to achieve the same reaction is related to the kinetic energy of the proton, we find:
\[ K_D = 2K_0 \]

### Final Answer:
The minimum kinetic energy \( K_D \) of the deuteron required to collide with a stationary proton and form a helium-3 nucleus is:
\[ K_D = 2K_0 \]